ECE215
Lecture – 5
• Second-Order Circuit (contd.)
Date: 18.08.2016
ECE215
Step Response of a Series RLC Circuit
• Apply KVL around the loop for t>0:
solution has two
components
• 𝑣𝑡 (𝑡) dies out with time and is of the form:
• 𝑣𝑠𝑠 (𝑡) is the final value of the capacitor voltage.
ECE215
Step Response of a Series RLC Circuit (contd.)
• the complete solutions are:
• The values of A1 and A2 are obtained from the initial conditions:
• Just to reiterate: v and i are the voltage across the capacitor and the current
through the inductor respectively.
• Above expressions help in finding v.
• once the capacitor voltage is 𝑣𝑐 = 𝑣 is known, we can determine 𝑖 = 𝐶 𝑑𝑣 𝑑𝑡
which is the same current through the capacitor, inductor, and resistor.
• the voltage across the resistor is 𝑣𝑅 = 𝑖𝑅 while the inductor voltage is 𝐿 𝑑𝑖 𝑑𝑡.
• Alternatively, the complete response for any variable 𝑥(𝑡)
can be found directly, because it has the general form:
ECE215
Example – 1
For this circuit, find v(t) for t >0.
• For t = 0-, the equivalent circuit is:
ECE215
Example – 1 (contd.)
• For t > 0, we have a series RLC circuit with a step input.
Underdamped
ECE215
Practice Example
A series RLC circuit is described by:
Find the response when L=0.5H, R=4Ω, and C=0.2F. Let i(0) = 1, di(0)/dt=0.
ECE215
Step Response of a Parallel RLC Circuit
• Apply KCL at the top node for t>0:
A1 and A2 to be determined
from 𝑖 0 and 𝑑𝑖(0) 𝑑𝑡.
𝑖 = 𝑖𝐿
• Alternatively, the complete response for any variable 𝑥(𝑡)
can be found directly, because it has the general form:
ECE215
Example – 2
find 𝑖(𝑡) for t > 0 in the following circuit:
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General Second-Order Circuits
• the series and parallel RLC circuits are the second-order circuits of greatest
interest, other second-order circuits including op amps are also useful.
• for a second-order circuit, we determine its step response 𝑥(𝑡)(which may be
voltage or current) by taking the following four steps:
1.
2.
3.
4.
We first determine the initial conditions 𝑥(0) and 𝑑𝑥(0) 𝑑𝑡 and the final
value 𝑥(∞).
We turn off the independent sources and find the form of the transient
response 𝑥𝑡 (𝑡) by applying KCL and KVL.
We obtain the steady-state response as: 𝑥𝑠𝑠 𝑡 = 𝑥(∞).
The total response is then found as the sum of the transient response and
steady-state response.
ECE215
Example – 3
In this circuit, find 𝑖(𝑡) for t > 0.
Example – 4
In the following circuit, the switch has been in position 1 for a long time but
moved to position 2 at t = 0, Find:
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Example – 6
For the op amp circuit, derive the
differential equation relating 𝑣𝑜 to 𝑣𝑠 .
Example – 5
In the following op amp circuit, determine
𝑣0 (𝑡) for 𝑡 > 0. Let 𝑣𝑖𝑛 𝑡 = 𝑢 𝑡 𝑉, 𝑅1 =
𝑅2 = 10 𝑘Ω, 𝐶1 = 𝐶2 = 100𝜇𝐹.
ECE215
Automobile Ignition System
• Earlier, considered the automobile
ignition system as a charging system. That
was only a part of the system.
• another part—the voltage generating
system.
• The 12-V source is due to the battery and
alternator.
• The 4Ω resistor represents the resistance
of the wiring.
• The ignition coil is modeled by the 8-mH
inductor.
• The 1𝜇𝐹 capacitor (known as the
condenser to auto mechanics) is in
parallel with the switch (known as the
breaking points or electronic ignition).
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Example – 7
An automobile airbag igniter is modeled by the following circuit. Determine the time it takes
the voltage across the igniter to reach its first peak after switching from A to B. Let R= 3Ω,
C=1/30pF, and L= 60mH.
Example – 8
A load is modeled as a 250-mH inductor in parallel with a 12Ω resistor. A capacitor
is needed to be connected to the load so that the network is critically damped at
60 Hz. Calculate the size of the capacitor.